Question: Captain Daniel has a ship, the H.M.S Crimson Lynx. The ship is two furlongs from the dread pirate Stephanie and her merciless band of thieves. If his ship hasn't already been hit, Captain Daniel has probability $\dfrac{3}{4}$ of hitting the pirate ship. If his ship has been hit, Captain Daniel will always miss. If her ship hasn't already been hit, dread pirate Stephanie has probability $\dfrac{1}{7}$ of hitting the Captain's ship. If her ship has been hit, dread pirate Stephanie will always miss. If the Captain and the pirate each shoot once, and the pirate shoots first, what is the probability that both the pirate and the Captain hit each other's ships?
Answer: The probability of event A happening, then event B, is the probability of event A happening times the probability of event B happening given that event A already happened. In this case, event A is the pirate hitting the Captain's ship and event B is the Captain hitting the pirate ship. The pirate fires first, so her ship can't be sunk before she fires her cannons. So, the probability of the pirate hitting the Captain's ship is $\dfrac{1}{7}$. If the pirate hit the Captain's ship, the Captain has no chance of firing back. So, the probability of the Captain hitting the pirate ship given the pirate hitting the Captain's ship is $0$. The probability that both the pirate and the Captain hit each other's ships is then the probability of the pirate hitting the Captain's ship times the probability of the Captain hitting the pirate ship given the pirate hitting the Captain's ship. This is $\dfrac{1}{7} \cdot 0 = 0$